Determine a série de Fourier de f no intervalo indicado.
f(x)=\left\{\begin{array}{rr}0, & -2<x<0 \\\ x, & 0 \leq x<1 \\\\ 1, & 1 \leq x<2\end{array}\right.

Passo 1

O enunciado pede a expansão em série de Fourier da função periódica de período 4 definida em [-2,2) por
f(x)= \begin{cases} 0, & -2<x<0,\\\ x, & 0\le x<1,\\\\ 1, & 1\le x<2. \end{cases}
Para resolver, usamos a forma
f(x)\sim \frac{a_0}{2}+\sum_{n=1}^{\infty}\bigl(a_n\cos\bigl(\tfrac{n\pi x}{2}\bigr)+b_n\sin\bigl(\tfrac{n\pi x}{2}\bigr)\bigr)
com
a_n=\frac{1}{2}\int_{-2}^{2}f(x)\cos\bigl(\tfrac{n\pi x}{2}\bigr)\,dx, \quad b_n=\frac{1}{2}\int_{-2}^{2}f(x)\sin\bigl(\tfrac{n\pi x}{2}\bigr)\,dx,
e
a_0=\frac{1}{2}\int_{-2}^{2}f(x)\,dx.

Passo 2

Cálculo de a_0.
a_0=\frac{1}{2}\Bigl(\int_{-2}^{0}0\,dx+\int_{0}^{1}x\,dx+\int_{1}^{2}1\,dx\Bigr) =\frac{1}{2}\Bigl(\frac{1^2}{2}+1\Bigr) =\frac{3}{4}.

Passo 3

Fórmula para a_n. Como f(x)=0 em [-2,0), escrevemos
a_n=\frac{1}{2}\Bigl(\int_{0}^{1}x\cos\bigl(\tfrac{n\pi x}{2}\bigr)\,dx +\int_{1}^{2}1\cdot\cos\bigl(\tfrac{n\pi x}{2}\bigr)\,dx\Bigr).
Chamando \alpha=\tfrac{n\pi}{2},
\int_{0}^{1}x\cos(\alpha x)\,dx =\frac{\cos(\alpha)-1}{\alpha^2}+\frac{\sin(\alpha)}{\alpha},
\int_{1}^{2}\cos(\alpha x)\,dx =\frac{\sin(2\alpha)-\sin(\alpha)}{\alpha}.
Logo
a_n=\frac{1}{2}\Bigl(\frac{\cos\alpha-1}{\alpha^2}+\frac{\sin\alpha}{\alpha} +\frac{\sin(2\alpha)-\sin\alpha}{\alpha}\Bigr) =\frac{1}{2}\Bigl(\frac{\cos\alpha-1}{\alpha^2}+\frac{\sin(2\alpha)}{\alpha}\Bigr).

Passo 4

Simplificação de a_n. Substituindo \alpha=\tfrac{n\pi}{2} e usando \sin(2\alpha)=\sin(n\pi)=0,
a_n =\frac12\;\frac{\cos\!\bigl(\tfrac{n\pi}{2}\bigr)-1}{\bigl(\tfrac{n\pi}{2}\bigr)^2} =\frac{2}{n^2\pi^2}\Bigl(\cos\bigl(\tfrac{n\pi}{2}\bigr)-1\Bigr).

Passo 5

Fórmula para b_n.
b_n=\frac{1}{2}\Bigl(\int_{0}^{1}x\sin(\alpha x)\,dx+\int_{1}^{2}\sin(\alpha x)\,dx\Bigr).
Com
\int_{0}^{1}x\sin(\alpha x)\,dx =-\frac{\cos(\alpha)}{\alpha}+\frac{\sin(\alpha)}{\alpha^2}, \quad \int_{1}^{2}\sin(\alpha x)\,dx =-\frac{\cos(2\alpha)-\cos(\alpha)}{\alpha},
obtém-se
b_n=\frac{1}{2}\Bigl(-\frac{\cos\alpha}{\alpha}+\frac{\sin\alpha}{\alpha^2} -\frac{\cos(2\alpha)-\cos\alpha}{\alpha}\Bigr) =\frac{1}{2}\Bigl(\frac{\sin\alpha}{\alpha^2}-\frac{\cos(2\alpha)}{\alpha}\Bigr).

Passo 6

Simplificação de b_n. Substituindo \alpha=\tfrac{n\pi}{2} e \cos(2\alpha)=(-1)^n,
\frac{\sin\alpha}{\alpha^2} =\frac{4\sin\!\bigl(\tfrac{n\pi}{2}\bigr)}{n^2\pi^2}, \quad -\frac{\cos(2\alpha)}{2\alpha} =-\frac{1}{2}\frac{(-1)^n}{\tfrac{n\pi}{2}} =\frac{(-1)^{\,n+1}}{n\pi}.
Logo
b_n =\frac{2\,\sin\!\bigl(\tfrac{n\pi}{2}\bigr)}{n^2\pi^2} +\frac{(-1)^{\,n+1}}{n\pi}.

Passo 7

Síntese da série. Com a_0=\tfrac{3}{4}, a_n e b_n obtidos, a expansão é
\boxed{% f(x)\sim \frac{3}{8} +\sum_{n=1}^{\infty}\Bigl[\frac{2}{n^2\pi^2}\bigl(\cos\bigl(\tfrac{n\pi}{2}\bigr)-1\bigr)\cos\bigl(\tfrac{n\pi x}{2}\bigr) +\Bigl(\frac{(-1)^{\,n+1}}{n\pi}+\frac{2\,\sin\bigl(\tfrac{n\pi}{2}\bigr)}{n^2\pi^2}\Bigr) \sin\bigl(\tfrac{n\pi x}{2}\bigr)\Bigr].}

Resposta

f(x)=\frac{3}{8} +\sum_{n=1}^{\infty}\Bigl[a_n\cos\bigl(\tfrac{n\pi x}{2}\bigr)+b_n\sin\bigl(\tfrac{n\pi x}{2}\bigr)\Bigr]
com
a_n=\frac{2}{n^2\pi^2}\Bigl(\cos\bigl(\tfrac{n\pi}{2}\bigr)-1\Bigr), \quad b_n=\frac{(-1)^{\,n+1}}{n\pi}+\frac{2\,\sin\bigl(\tfrac{n\pi}{2}\bigr)}{n^2\pi^2}.

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